Problem 54 - Calculus and its applications

Question

Below are the steps in the simplification of the difference quotient for \(f(x)=\sqrt{x}\) (see Example 8). Provide a brief justification for each step. \(\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\) a) \(=\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\) b) \(=\frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})}\) c) \(=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\) d) \(=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\) e) \(=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Step-by-Step Solution

Answer

\(\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Step 1: Understanding the Original Expression

The given expression \( \frac{f(x+h)-f(x)}{h} \) simplifies to \( \frac{\sqrt{x+h}-\sqrt{x}}{h} \). This is known as the difference quotient and helps in finding the derivative of \( f(x) = \sqrt{x} \).

Step 2: Rationalizing the Numerator

To simplify \( \frac{\sqrt{x+h}-\sqrt{x}}{h} \), we multiply both the numerator and the denominator by the conjugate \( \sqrt{x+h}+\sqrt{x} \). This is step (a) \( = \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right) \).

Step 3: Applying Difference of Squares

Use the difference of squares: \((a-b)(a+b) = a^2 - b^2\). When multiplying \( \sqrt{x+h}-\sqrt{x} \) by its conjugate, we get \( x+h-x = h \), leading to \( \frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})} \) as in step (b).

Step 4: Simplifying the Numerator

The expression simplifies to \( \frac{h}{h(\sqrt{x+h}+\sqrt{x})} \), observed in steps (c) and (d) after cancelling out the \( x \) terms and simplifying \( x+h-x \) to \( h \).

Step 5: Cancelling the Common Factor

In step (d), the term \( h \) in the numerator and denominator cancel each other out, resulting in \( \frac{1}{\sqrt{x+h}+\sqrt{x}} \).

Step 6: Simplified Form of the Difference Quotient

The final simplified form of the difference quotient, as given in step (e), is \( \frac{1}{\sqrt{x+h}+\sqrt{x}} \). This expression is crucial for finding the derivative at a point.

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus that helps us find the derivative of a function. In simple terms, it's a way to measure how a function changes as its input changes. Consider the function we are dealing with here: - Given by: \[ f(x) = \sqrt{x} \] - We examine the expression for change: \[ \frac{f(x+h)-f(x)}{h} \] - For our function, it becomes: \[ \frac{\sqrt{x+h} - \sqrt{x}}{h} \] The difference quotient is how we prepare to find limits and derivatives. The process sets up a scenario where the division reflects a rate of change over a small interval, with "h" being the difference between points. As h approaches zero, this becomes the derivative of the function.

Rationalizing the Numerator

Rationalizing the numerator is a technique used to simplify expressions with square roots, especially useful here for differentiating functions like square roots. Let's explore how it is applied: When we encounter the expression: \[ \frac{\sqrt{x+h} - \sqrt{x}}{h} \] - Multiplying both the numerator and denominator by the conjugate of the numerator clears the square roots. - The conjugate involves switching the minus sign to a plus sign: \[ (\sqrt{x+h} + \sqrt{x}) \] - Performing this multiplication looks like: \[ \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \] The benefit? With this clever trick, the difference of squares rule simplifies the numerator significantly because: - √a² - √b² turns into just a – b, which is much easier to handle.

Simplification of Expressions

Simplification of expressions is key in making the math manageable. By applying algebra and logic, we reduce the complexity of expressions. Here are some important steps in this process: When dealing with our expression after rationalizing: - Start with: \[ h = x+h-x \] This clears out unnecessary parts, leaving us with the simplified form: \[ \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \] - Notice the terms in the numerator and denominator are the same, allowing us to "cancel" h: \[ \frac{1}{\sqrt{x+h} + \sqrt{x}} \] - This final simplification is more efficient and reveals the derivative's format. Each cancellation and simplification step brings clarity, making it possible to find derivatives more easily, which is a significant goal in calculus.